Let `sin x + sin y = cos x + cos y` for all `x, y in R`. What is `tan(x/2 + y/2)` equal to?

Options

A. `1`
B. `2`
C. `sqrt(2)`
D. `2sqrt(2)`

Answer

To solve the equation \( \sin x + \sin y = \cos x + \cos y \) for all \( x, y \in \mathbb{R} \) and find the value of \( \tan\left(\frac{x}{2} + \frac{y}{2}\right) \), we can follow these steps:

### Step 1: Use the sum-to-product identities
We can rewrite the left-hand side and right-hand side using the sum-to-product identities:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]

### Step 2: Set the two expressions equal
From the given equation, we have:
\[ 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]

### Step 3: Cancel the common factor
Assuming \( \cos\left(\frac{x-y}{2}\right) \neq 0 \), we can divide both sides by \( 2 \cos\left(\frac{x-y}{2}\right) \):
\[ \sin\left(\frac{x+y}{2}\right) = \cos\left(\frac{x+y}{2}\right) \]

### Step 4: Use the identity for tangent
The equation \( \sin\left(\frac{x+y}{2}\right) = \cos\left(\frac{x+y}{2}\right) \) implies that:
\[ \tan\left(\frac{x+y}{2}\right) = 1 \]

### Step 5: Solve for \( \frac{x+y}{2} \)
This means:
\[ \frac{x+y}{2} = \frac{\pi}{4} + n\pi \quad \text{for some integer } n \]

### Step 6: Find \( \tan\left(\frac{x}{2} + \frac{y}{2}\right) \)
Now, we want to find \( \tan\left(\frac{x}{2} + \frac{y}{2}\right) \):
\[ \frac{x}{2} + \frac{y}{2} = \frac{x+y}{2} \]
Thus:
\[ \tan\left(\frac{x}{2} + \frac{y}{2}\right) = \tan\left(\frac{\pi}{4} + n\pi\right) \]

Since the tangent function has a period of \( \pi \):
\[ \tan\left(\frac{\pi}{4} + n\pi\right) = 1 \]

### Final Answer
Therefore, the value of \( \tan\left(\frac{x}{2} + \frac{y}{2}\right) \) is:
\[ \boxed{1} \]